Download 111 Exam II Outline

CHEMISTRY 111 LECTURE
EXAM II Material
REVIEW
A. Trends
1. Atomic Radius [Atom Size]
a. As the number of shells increases, the radius size increases
b. As you go left to right across a period, the radius size decreases
2. Electron Affinity The amount of energy released or absorbed when an electron is added to an
atom to form a (-) ion [anion], in gas phase.
a. As you go left to right across a period, the electron affinity increases.
b. As you go down a group (top to bottom), the electron affinity decreases.
3. Ionization Energy
The energy required to remove an electron from a neutral atom (in gas phase).
a. As you go left to right across a period, the ionization energy increases.
b. As you go down a group (top to bottom), the ionization energy decreases.
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4 Ion Size - is the measure of the electron cloud around the nucleus.
a. Cations
b. Anions
B. Isoelectronic particles - species with the same number of electrons
Problem: Arrange the ions Se2-, Br-, Rb+, and Sr+2 in decreasing size
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The attractive interaction between two atoms or ions
I. Types: 1. Ionic Bond- Cations (+ charged) and Anions (- charged) are held together by the
attractive force of their (+) and (-) charges  Electrostatic force.
2. Metallic Bonds
3. Macro molecular crystals
4. Covalent Bonds- Results from the sharing of a pair of electrons between two atoms.
II. Valence electrons (High energy electrons)
The electrons in the outermost shell (energy level). Valence electrons are involved in reactions.
(Rem: # valence e- = the group number for the "A" subgroup elements)
ex.
III. Ionic Bonds- The attractive force between a cation (+ ion) and anion (- ion).
Atoms lose or/ gain electrons to obtain an octet.
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IV. Covalent Bonds
A. Bond Energy - The average energy required for the dissociation of a bond
B. Bond Length - The average distance between the two nuclei of covalently bonded atoms.
C. Drawing electron dot structures
HOW TO:
1.
Write e- dot structure for the individual atoms.
2.
a) Add together the number of valence electrons for all the atoms
(If it is an ion, you must add or subtract electrons accordingly)
b) Divide the total number of e- by 2: This will give you the number of e- pairs available
for bonding.
3.
Determine which is the central atom
a. The least represented atom that is not H
b. Usually, the first atom in the chemical
formula that is not H.
4.
Arrange atoms symmetrically around the central atom.
5.
Draw a single line (or 2 dots) between the central and outer atoms.
6.
From the total number of valence electrons subtract 2 electron for each bond made.
7.
Attempt to place the remaining electron pairs around the outer atoms to make an octet
or duet (for H)
8.
Additional electrons are placed on the central atom
9.
If the central atom still has less than an octet ; then, a double or triple bond must be formed.
Do not use a double or triple bond unless you have to!
, a general rule(a help)
Examples.
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B. Specific Electron Dot Cases:
1. Ions:
2. Oxy Acids
3. Carbon chains
D. Exceptions to the Octet Rule
1. Electron deficient molecules: Molecules where the central atom does not have an octet. Usually a group IIIA atom
Example: BCl3
2. Expanded valence shell: Molecules where the central atom has more than 8 valence around the central atom. The
central atom would belong to the 3rd, 4th,5th,6th,or 7thperiod.
Example: SF6
3. Molecules with an odd number of electrons: There are an odd number of valece electrons in the molecule
Example:
NO2
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Practice:
SO2
IF5
NO3-
HBrO
CH3COCH3
PF5
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V. Intra and Intermolecular (particle) forces
A.
B. Intramolecular (particle) forces
C. Intermolecular (particle) forces
The attractive forces between particles.
Types
1. Dipole-Dipole interaction:
Dipole - dipole interactions are electrostatic attractions between polar molecules
2. Hydrogen bonds:
A hydrogen bond is a relatively strong dipole-dipole attractive force between a hydrogen atom
and a pair of nonbonding electrons on a F,O, or N atom
3. London forces
The attraction between atoms and nonpolar molecules. London forces are very weak electrostatic
forces of attraction between molecules with "temporary" dipoles.
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VI. Electronegativity and bonding
A. Electronegativity - The measure of the attractive force that an atom for its shared electrons. In general,
electronegativity increases left to right and bottom to top on the periodic table.
B. Electronegativities and bond polarity
1. Covalent Bonds
a. Non polar covalent bonds - differences in the electronegativities is
< 0.4
b. Polar covalent bonds - differences in the electronegativities is between 0.5 -1.7
2. Ionic bonds
Differences in electronegativities is
> 1.7
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Excersizes
1.
What is the major type of intermolecular forces for the following
I2
PCl3(pyrimidal)
H2S (bent)
CO
CO2 (linear)
CCl4 (tetrahedral)
CH3CH2NH2
CH3OH (tetrahedral)
CH2O (Trigonal Planar)
CH4(Tetrahedral)
2.
Which has the strongest intermolecular forces?
I2
CH3CH3
or
Br2
or
or
Cl2
CH3CH2CH2CH2CH2CH2CH2CH3
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or
CH3CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH2CH3
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A. Types
1.
Crystalline Solid
Crystalline solids have a highly ordered arrangement of particles (ions, atoms, and molecules)
2. Amorphous Solid
Amorphous solids have considerable disorder in their structure.
B. Crystalline Solids
1. Crystalline lattice
A three dimensional array of lattice points in a pattern that defines a crystal.
2. Unit Cell
The Unit cell is the basic repeating unit of the lattice.
3. Coordination Number
The coordination number of a particle in a crystal is thenumber of nearest neighbors .
4. Lattice Points
The points in a lattice occupied by atoms, ions or molecules.
5. Kinds of Lattice
.
b
a


c

Simple cubic
a=b=c
 =  =  = 90
Monoclinic
a≠b≠c
 =  = 90,  ≠ 90
Tetragonal
a=b≠c
 =  =  = 90
Orthorhombic
a≠b≠c
 =  =  = 90
Triclinic
a≠b≠c
 ≠  ≠  ≠ 90
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Rhombohedral
a=b=c
 =  =  ≠ 90
Hexagonal
a=b≠c
 =  = 90,  = 120
5. Geometry
6.
Simple Cubic
a. Calculate the volume of the unit cell
b. Calculate the volume of the spheres
c. Calculate the percent occupied (packing efficiency)
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6.
Body Centered Cubic
a. Calculate the volume of the unit cell
b. Calculate the volume of the spheres
c. Calculate the percent occupied (packing efficiency)
7. Face Centered Cubic
a. Calculate the volume of the unit cell
b. Calculate the volume of the spheres
c. Calculate the percent occupied (packing efficiency)
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Problems:
1. Molybdem has an atomic radius of 136 pm and crystallizes in a body centered cubic system.
a. What is the length of the edge of a unit cell.
b. Calculate the density of molybdem
2. Aluminum crystallizes in the face-centered cubic system, and the edge of a unit cell is 287.5 pm. Calculate
atomic radius of Al.
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the
8.
Packing Efficiency
a. Closest cubic packing
b. Hexagonal
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I. PROPERITIES OF GASES
A. Gas particles are far apart from each other - there is no attraction between particles
B. Gases have an indefinite shape.
C. Gases have a low density
D. Gases are very compressible
E. Gases exert pressure equally in all directions on the walls of a container.
F Gases have a high flow rate. Gases mix spontaneously and completely with one or more other
gases.
II. PROPERITIES OF SOLIDS
A. A Solid contain particles which are very close to each other-there are very large attractive forces between
particles
B. Solids have a definite shape. Solids maintain its shape regardless of the container they are in.
C. Solids in general have a high density
D. Solids are not compressible (or negligibility)
`
E. Solids do not flow
III. PROPERITIES OF LIQUIDS
A. Liquids contain particles which are (somewhat) close to each other - there are attractive forces
between particles
B. Liquids have a definite shape. Liquids maintain the shape of the bottom of the container.
C. Liquids in general have a medium density
D. Liquids are not compressible (or negligibility)
E. Liquids have a medium flow rate
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I. Properties of gases
II. Measurements
A. Pressure =
force
Unit area
1. Conversions:
`
1 atm= 760 mm Hg = 760 torr (exactly)
1.013 x 105 Pa= 1 atm = 14.68 psi
2. Barometer
3. Manometer
B. Temperature - Kelvin
K = ºC + 273
C. Volume
1. The volume of a gas is the volume of the container it occupies.
2. Units: liters or milliliters
III. KINETIC MOLECULAR THEORY
A. Gases are composed of such extremely tiny atoms or molecules that are widely separated by empty space.
B. Gas particles move in a random,rapid, and continuous motion, thus has kinetic energy.
C. Gas particles moves so rapidly and are so far apart the there is essentially no force of attraction between the
particles.
D. Particles collide frequently wtih each other and with the walls of the container, the collisions are perfectly
"elastic" - (No net loss of energy as a result of a collision)
IV. AVERAGE KINETIC ENERGY
The average kinetic energy (energy of motion) of the gas particles are directly proportional to its absolute T º (Kelvin)
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V. RELATIONSHIP BETWEEN oT, VOLUME, AND PRESSURE.
A. Boyle's law
P&V
As the pressure increases the volume decreases in the same proportion.
ºT & V
B Charles's law
As the temperature (Kelvin) is increased the volume is increased proportionally.
C Gay-Lussac's Law
When temperature (K) increases pressure increases proportionally.
D. COMBINED GAS LAW
P,V, and oT varying. Assume that the mass is constant.
A certain mass of gas occupies 5.50 L at 34ºC and 655 mm Hg. What will its volume in liters be if it is cooled to
10.0oC and its pressure remains the same.
E. GAY-LUSSAC'S LAW OF COMBINING VOLUMES
At the same oT and Pressure, the volumes of gases that combine in a chemical reaction are in the ratio of small
whole numbers.
F MOLAR GAS VOLUME; AVOGARDO'S HYPOTHESIS
At the same temperature and pressure the same number of moles of different gases have the same volume. The
Molar Volume is the volume of one mole of any gas at a given oT & P. [STP]
Standard Temperature and Pressure = [STP]:
22.4 L
1 mole of gas
At:
273 K and 1 atm (760 torr)
The density of an unknown gas is 1.43 g/L at 0ºC and 760 torr. What is the molar mass of the unknown gas?
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G. IDEAL GAS EQUATION:
Derivation:
KNOW:
PV=nRT
Where: n = moles of gas
R = 0.0821 L-atm
mole-K
1. What volume in liters will be occupied by 6.00 mol carbon dioxide gas at 105 mm Hg and 28ºC?
WHEN TO USE:
1. PV = nRT
22.4 L
2. mole
3.
at STP
P1V1
P2V2
=
T1
T2
H. DALTON'S LAW OF PARTIAL PRESSURES; Mixtures of gases
The total pressure of a mixture of gases is equal to the sum of the partial pressures exerted by each gas.
Ptotal = P1 + P2 +P3 +.....
Example: The total pressure in a 1.00 liter container is 725 mm Hg. The container contains water vapor and
nitrogen gas.
If the partial pressure of the water vapor is 225 mm Hg, what is the partial pressure of the nitrogen gas.
Ptotal = PN + PH O
2
2
I. MOLE FRACTIONS; Mixtures of gases
The mole fraction of a component is the fraction of moles of that component of the total moles of the gas
mixture.
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VI GAS STOICHIOMETRY
Certain chemical reactions involve gas as a reactant or product. For these types of reactions, the stoichiometric
calculations involve the use of:
1) PV=nRT
L
2) 22.4 mole at STP
3) Molar volumes
The general stoichiometric scheme
Vol. of known (gas)
Vol. of unknown (gas)
PV=nRt or 22.4 L/mole (at STP) or molar volumes
g of Known
Moles of Known
Moles of UNK.
g. of UNK.
Molarity (mol/L)
Vol. of Known
(liters)
Vol. of UNK.
(liters)
Problems:
1. How many liters of ammonia gas can be produced by the reaction of 735 ml hydrogen gas with an excess nitrogen
gas at 425 oC and 135 atm?
Nitrogen + hydrogen --> ammonia
Ans.=0.490 L
2. How many liters of carbon dioxide gas at 0 oC and 1 atm are produced by the complete combustion of 60.0 mol of
liquid glucose, C6H12O6?
Ans. = 9.10 x 103 CO2
3. How many liters of the air pollutant NO(g) could be produced at 985 oC and a pressure of 30.0 atm by the reaction of
oxygen gas with 455 g of nitrogen gas.
Ans. = 112 L NO
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VII GAS PROBLEMS
1. A 655 ml gas cylinder filled with oxygen gas at a pressure of 95 atm and at 26.0 °C was used by a scuba diver. The
pressure after it was used was 85 atm. How many moles of oxygen gas were used by the diver?
Ans = 0.2 mol O2
2. A flask contained 1.017 mol of carbon dioxide. The gas exerted a pressure of 925 mm Hg at a temperature of 28 °C.
When an additional 0.250 mole of Carbon dioxide was added to the flask the temperature increased to 35°C. What is
the new pressure in the flask?
Ans.= 1.56 atm CO2
3. A sample of an unknown gaseous hydrocarbon had a density of 1.56 g/L at 25.0 °C AND 1.33 atm. Calculate the molar
mass of the gas.
Ans. = 28.7 g/mol
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4. A container with only He had a pressure of 544 torr at a temperature of 35 °C. When 0.810 g of Ne is added to this
container, the pressure increases to 959 torr. Calculate the grams of He in the container.
Ans. = 0.212 g He
5. 6.53 x 1028 molecules of Oxygen occupy 15.00 liters. What is the volume occupied by 66.5 g of carbon dioxide under
the same conditions?
Ans. = 2.10 x 10-4 L CO2
6. A mixture containing 1.22 g Xe and 0.675 g NO2 exerts a pressure of 1.44 atm. What is the partial pressure of NO2?
Ans. = 0.883 atm NO2
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7. The complete combustion of 0.500 g of hydrocarbon, containing only C and H, produced 0.771 L of CO2 at STP and
0.755 g of water. In another experiment, 0.218 g of sample occupied 185 ml at 23 °C and 374 mm Hg. What is the
molecular formula of the compound?
Ans. = C4H10
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VIII MOLECULAR SPEEDS; DIFFUSION AND EFFUSION
A. MOLECULAR SPEEDS
# of
molecules
molecular speed
B. DIFFUSION AND EFFUSION
Diffusion is the ability of two or more gases to spontaneously mix until it becomes a uniform,
homogeneous mixture.
Effusion is the process by which gas particles flows thru a very small hole from a container of high
pressure to a lower pressure.
Graham's Law of Effusion - The rate of effusion of a gas is inversely proportional to
it's size [ Molar Mass ] at constant temperature and pressure.
rA
MB
rB =
MA
Problem #1: What is the rate of effusion for H2 if 15.00 ml carbon dioxide of CO2 takes 4.55 sec to effuse out of a
container?
Problem #2: What is the molar mass of gas X if it effuses 0.876 times as rapidly as N 2(g)?
Page 24
IX REAL GASES
Gas laws describe the behavior of an ideal or "perfect" gas - a gas described by the kinetic molecular theory.
Under normal conditions of typical pressure and temperature, gases follow the ideal gas laws fairly closely.
At low temperature and/or high pressures gases deviate from the ideal gas laws.
A. Intermolecular forces of attraction
B. Molecular volume
Page 25
IONIC
PARTICLES AT
LATTICE
POINTS
cations and anions
[positive and
negative ions]
BONDING
BETWEEN
PARTICLES
(attractive forces)
electrostatic
attractions
[ionic bond]
PROPERITIES
high M.P.
high B.P.
nonconductors
brittle
hard/brittle
EXAMPLES
NaCl
KBr
LiNO3
MOLECULAR
polar or nonpolar
covalent molecules
intermolecular
forces
dipole-dipole,
hydrogen bond,
london forces
moderate to low
M.P.
moderate to low
B.P.
nonconductors or
semi conductors
soft
H2 O
SO2
NH3
C2H5OH
Page 26
NETWORK
COVALENT
METALLIC
atoms
metallic cations
[positive metallic
ions]
covalent bonds
metallic bonds
between positive
metal ions and
mobile electrons
high M.P.
high B.P.
nonconductors
brittle
hard
variable M.P.
variable B.P.
malleable and
ductile
C (diamond)
SiC
SiO2
Au
Cu
Fe
Na
Hg
BACKGROUND:
I. Law of Hess-Law of constant heat summation
∆ H rxn = ∆Hproducts - ∆Hreactants
A. Enthalphy, ∆H
REACTANTS
PROCUCTS
PRODUCTS
REACTANTS
B. Energy diagram
C (graphite) + O2 (g)
ΔH = -110.5KJ
CO(g) + 1/2
O
2
ΔH = -393.5 KJ
ΔH = -283.0 KJ
C O 2 (g)
C. Energy Equations:
1. Additon of equations
C (graphite) + O2(g)
CO (g) + 1/2 O2 (g)
CO(g) + 1/2 O2
CO2
_________________________________________________________
Page 27
∆H = -110.5 KJ
∆H = -283.0 KJ
2. Multiplying:
HI(g)

1/2 H2(g) + 1/2 I2(s)
∆H = -25.9 KJ
3. Reversing:
1/2 H2(g) + 1/2 I2(s)

HI(g)
∆H = +25.9 KJ
4. Problem:
Find ∆H for 2 NH3 (g) + 3 N2O(g)
Given:
4 NH3 (g) + 3 O2(g)
N2O(g) + H2(g)
H2O (l)




4N2(g) + 3 H2O (l)
2 N2(g) + 6 H2O (l)
N2(g) + H2O (l)
H2(g) + 1/2 O2(g)
Page 28
∆H = -1531 KJ
∆H = -367.4 KJ
∆H = +285.9 KJ
II BORN-HABER CYCLE
The Born- Haber cycle uses the law of Hess to determine the Lattice Energy. The lattice energy is the enthalphy
change, ∆H, associated when gaseous cations and anions from a crystal:
Na+(g) + Cl-(g)  NaCl(s) ∆H = - 788KJ
Since heat is always evolved in these processes, all lattice energies have a negative sign (Energy is lost). Lattice
enrgies cannot be determined directly; therefore, the law of Hess is applied to indirectly determine the lattice
energy:
To prepare 1 mole of NaCl, for example, the following steps must be taken:
A. Crystalline Na metal is sublimed into gaseous Na:
∆Hsublimation = +108 KJ
B. Gaseous sodium atoms are ionized into gaseous sodium ions. The amount of energy required for 1 mole
of Na is the first ionization energy of sodium
∆Hionization = +496 KJ
*Note some cations will require the second ionization energy step:
C. Gaseous Cl2 molecules are dissociated. (Bonds are broken). This the enthalpy change when 1 mole of
Cl2 molecules are dissociated.
∆Hdissociation = +243 KJ
D. Electrons are added to the gaseous chlorine atoms.
electron affinity of chlorine
The enthalpy change per mole of Cl(g) is the first
∆Helectron affinity = -349 KJ
*Note some anions will require the second electron affinity energy step:
E. In this step, gaseous Na+ and Cl- condense into one mole of crystalline sodium chloride. The lattice
energy is the amount of energy released per mole of NaCl.
∆Hlattice energy = -788 KJ
F. Once steps A-E are added, the sum will give you the ∆Hformation:
Na(s) + 1/2 Cl2 (g)

NaCl(s)
Page 29
∆Hformation = -411 KJ
Comparision between Lattice energy and ∆Hformation:
Problem: Calculate the Lattice energy for:
Li+ + F-  LiF
Given:
∆Hsublimation = +155.2 KJ
∆Hionization = +520. KJ
∆Hdissociation = +150.6 KJ (Bond energy)
∆Helectron affinity = -333 KJ
∆Hformation = -594.1 KJ
Page 30
∆Hlattice energy = ?
Problem: Calculate the Lattice energy for:
Mg2+ + 2Cl-

MgCl2
∆Hlattice energy = ?
Given:
∆Hsublimation = +120. KJ
∆Hionization : the first and second ionization energy of Mg is:
∆Hdissociation = +248 KJ (Bond energy)
∆Helectron affinity = -368 KJ
∆Hformation = -698 KJ
Page 31
756 KJ/mole and 1490. KJ/mole, respectively.
A. Comparison of the three physical states of matter
B. EVAPORATION OF LIQUIDS
Evaporation is the change from the liquid state to the gaseous or vapor state.
Liquid + heat  Vapor
1. Effect of intermolecular forces on evaporation
2. Effect of temperature on evaporation
3. Effect of surface area on surface area
Page 32
4. Effect of evaporation on the average kinetic energy
Number
of
Molecules
Kinetic Energy
C. VAPOR PRESSURE OF LIQUIDS
The vapor pressure of a liquid is the pressure exerted by the vapor above a liquid.
Dynamic equilibrium : rate of evaporation = rate of condensation
1. Effect of intermolecular forces on vapor pressure
2. Effect of temperature on vapor pressure
Vapor
Pressure
Temperature, °C
*Note: The vapor pressure of a liquid is independent of the volume of the container, provided that there is some
liquid present so that equilibrium is established
Page 33
D. BOILING POINT
The boiling point of a liquid is the temperature at which a liquid is changed to a gas within the liquid (bubbles formed
underneath the surface). The boiling point is the temperature at which the vapor pressure equals the atmospheric
pressure (external pressure)
1. The normal boiling point is the temperature at which the vapor pressure equals the atmospheric pressure of 1 atm.
2. Effect of intermolecular forces on the boiling point
Page 34
A. HEAT CAPACITY [Specific heat]
The amount of heat required to raise the temperature of 1 g of a substance exactly 1oC.
Example: How many degrees Celsius will the temperature rise if 25 g ether absorbs 160. cal of energy.
0.529 cal
Specific heat ether =
g °C
B. ENERGY AND CHANGE OF STATE
Energy (as heat) is either lost or absorbed when a substance changes its state
Solid
Liquid
Liquid
Gas
Gas
Solid
.
C. HEAT OF VAPORIZATION- The quantity of heat needed to convert a liquid at its boiling point to the gaseous
state.
Ex.
∆ Hvap =
2.26 KJ
g
D. HEAT OF FUSION - The quantity of heat needed to convert a solid at its melting point to the liquid state.
Ex.
∆ Hfus =
3.35J
g
Page 35
Problem.
How much energy [Heat in kilojoules] is needed to convert 500.0 g of ice at -15.0oC to steam at 105.0 oC?
Hfusion= 335 J
Hvap = 2.26 KJ
g
g
Specific heat of ice
=
2.10 J
g oC
Specific heat of water =
4.18 J
g oC
Specific heat of steam =
2.0 J
g oC
GRAPH:
T
E
M
P
E
R
A
T
U
R
E
GAS
LIQUID  GAS
D
100 C
E
LIQUID
SOLID  LIQUID
B
0C
C
SOLID
A
HEAT ADDED
Page 36
Calculations:
AB
HEATING A SOLID
BC
SOLID  LIQUID
CD
HEATING A LIQUID
DE
LIQUID  GAS
EF
HEATING A GAS
TOTAL HEAT ADDED
E. CRITICAL TEMPERATURE, CRITICAL PRESSURE, AND CRITICAL POINT
To convert a gas to a liquid, you must:
a. Lower the temperature
and/or
b. Increase the pressure
1. The Critical temperature of a gas is the highest temperature above which a substance can only exist as a gas.
Above its critical temperature, no increase in pressure will convert the gas into a liquid.
2. The Critical pressure is the minimum vapor pressure at the critical temperature. The minimum pressure required to
convert the gas into a liquid is its critical temperature.
Page 37
F. PHASE DIAGRAMS
A phase diagram is a graph of pressure versus temperature that shows the region of stability for each of the physical
states. The three line segments show the combinations of pressure and temperature at which any two phases exist in
equilibrium.
1. The line that separates the gas from the liquid is the vapor pressure curve
2. The Triple point is the pressure and temperature at which solid, liquid, and gas all coexist at the same time.
3. The liquid-gas equilibrium line ends at the critical point. Above the critical temperature only a gas exists.
Page 38
PHASE DIAGRAM FOR H2O
Page 39
PHASE DIAGRAM FOR CO2
Page 40